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Re: Bhai's Probability 10 [#permalink]
18 Nov 2003, 02:24

Bhai wrote:

A deck of ordinary cards is shued and 13 cards are dealt. What is the probability that the last card dealt is an ace?

The answer is 1/13. Since we have no information about any of the cards, the probability of ANY CARD being an ace is 1/13. _________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

Re: Bhai's Probability 10 [#permalink]
19 Nov 2003, 16:16

Total outcomes = 52P13
Possible outcomes= 4.51P12 (13th card can be an ace in 4 ways)
probability = 4/52=1/13

Akamaibrah- If cards had sets of 3 Aces and not 4, then the answer would have been 3/52, therefore I am not too sure if we can solve this problem in the method you suggested. Please clarify if I am wrong.

AkamaiBrah wrote:

Bhai wrote:

A deck of ordinary cards is shued and 13 cards are dealt. What is the probability that the last card dealt is an ace?

The answer is 1/13. Since we have no information about any of the cards, the probability of ANY CARD being an ace is 1/13.

Re: Bhai's Probability 10 [#permalink]
21 Jan 2004, 07:52

AkamaiBrah wrote:

Bhai wrote:

A deck of ordinary cards is shued and 13 cards are dealt. What is the probability that the last card dealt is an ace?

The answer is 1/13. Since we have no information about any of the cards, the probability of ANY CARD being an ace is 1/13.

Akmai,

I hope you can help me out with the question I have. Wouldn't this be the solution only if we assume that the each of the first 12 cards is put back in the deck?

If we assume that the cards are not put back in the deck , then I think we need to consider the followig probabilities:

(1) Probability that there is EXACTLY one Ace in the first 12 cards and the 13th card is Ace again = [12C1*(4/52)(48/52)^11] *3/39

(2) Probability that there are EXACTLY 2 aces in the first 12 cards and the 13th card is Ace again = [12C2 * (4/52)^2 * (48/52)^10] * 2/39

Similarly, the probability with (exactly 3 and 13th card ace) and with (exactly 4 and the 13 card ace) can be found.

The required probability would be the total of all the probability found above.

Re: Bhai's Probability 10 [#permalink]
03 Feb 2004, 07:13

gmatblast wrote:

AkamaiBrah wrote:

Bhai wrote:

A deck of ordinary cards is shued and 13 cards are dealt. What is the probability that the last card dealt is an ace?

The answer is 1/13. Since we have no information about any of the cards, the probability of ANY CARD being an ace is 1/13.

Akmai,

I hope you can help me out with the question I have. Wouldn't this be the solution only if we assume that the each of the first 12 cards is put back in the deck?

If we assume that the cards are not put back in the deck , then I think we need to consider the followig probabilities:

(1) Probability that there is EXACTLY one Ace in the first 12 cards and the 13th card is Ace again = [12C1*(4/52)(48/52)^11] *3/39

(2) Probability that there are EXACTLY 2 aces in the first 12 cards and the 13th card is Ace again = [12C2 * (4/52)^2 * (48/52)^10] * 2/39

Similarly, the probability with (exactly 3 and 13th card ace) and with (exactly 4 and the 13 card ace) can be found.

The required probability would be the total of all the probability found above.

Please clarify why this approach would be wrong.

If you add up all of the probabilities of ALL of the contigencies, the sum total will be 1/13.

Think of this: Suppose you deal 1 card off the deck. The probabilty is clearly 1/13. Now reshuffle and deal 13 cards off the top of the deck and put the cards back on the deck with the 13th card on top. Can you think of any reason why this card should have any probability other than 1/13? _________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

Re: Bhai's Probability 10 [#permalink]
03 Feb 2004, 07:55

AkamaiBrah wrote:

gmatblast wrote:

AkamaiBrah wrote:

Bhai wrote:

A deck of ordinary cards is shued and 13 cards are dealt. What is the probability that the last card dealt is an ace?

The answer is 1/13. Since we have no information about any of the cards, the probability of ANY CARD being an ace is 1/13.

Akmai,

I hope you can help me out with the question I have. Wouldn't this be the solution only if we assume that the each of the first 12 cards is put back in the deck?

If we assume that the cards are not put back in the deck , then I think we need to consider the followig probabilities:

(1) Probability that there is EXACTLY one Ace in the first 12 cards and the 13th card is Ace again = [12C1*(4/52)(48/52)^11] *3/39

(2) Probability that there are EXACTLY 2 aces in the first 12 cards and the 13th card is Ace again = [12C2 * (4/52)^2 * (48/52)^10] * 2/39

Similarly, the probability with (exactly 3 and 13th card ace) and with (exactly 4 and the 13 card ace) can be found.

The required probability would be the total of all the probability found above.

Please clarify why this approach would be wrong.

If you add up all of the probabilities of ALL of the contigencies, the sum total will be 1/13.

Think of this: Suppose you deal 1 card off the deck. The probabilty is clearly 1/13. Now reshuffle and deal 13 cards off the top of the deck and put the cards back on the deck with the 13th card on top. Can you think of any reason why this card should have any probability other than 1/13?

AkamaiBrah,

Thanks for the explanation. It now makes sense. It is amazing to see that the probability of all the alternatives will add up to 1/13. But I agree that in this example, it is important to look at the problem in the way you did.

gmatclubot

Re: Bhai's Probability 10
[#permalink]
03 Feb 2004, 07:55

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...